A) \[{{d}^{2}}+{{(2b-3c)}^{2}}=0\]
B) \[{{d}^{2}}+{{(3b-2c)}^{2}}=0\]
C) \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\]
D) \[{{d}^{2}}+{{(3b-2c)}^{2}}=0\]
Correct Answer: C
Solution :
Given parabolas are \[{{y}^{2}}=4ax\] ?..(i) and \[{{x}^{2}}=4ay\] On putting the value of y from Eq. (ii) in Eq. (i), we get \[\frac{{{x}^{4}}}{16{{a}^{2}}}=4ax\] \[\Rightarrow \] \[x({{x}^{3}}-64{{a}^{3}})=0\] \[\Rightarrow \] \[x=0,\,4a\] From, Eq. (ii) we get, \[y=0,\,4a.\] Let \[A=(0,0),\,B=(4a,4a)\] Since, given line \[2bx+3cy+4d=0\] ?.(iii) Passes through intersection points A and B. On putting \[x=0,\,y=0\] in Eq. (iii), we get \[d=0\] On putting \[x=4a,\,y=4a\] and \[d=0\] in Eq. (iii), we get \[8ab+12ac=0\] \[\Rightarrow \] \[2b+3c=0\] \[(\because \,\,a\ne 0)\] Now, obviously \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\]
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